Properties of Triangle — Page 5 | JEE Mathematics

Q41

The sides of a triangle are

\sin \alpha ,\,\cos \alpha

and

\sqrt {1 + \sin \alpha \cos \alpha }

for some

0 < \alpha < {\pi \over 2}

. Then the greatest angle of the triangle is :

A

{150^ \circ }

B

{90^ \circ }

C

{120^ \circ }

D

{60^ \circ }

Correct Answer

Option C

Solution

Let

a = \sin \alpha ,b = \cos \alpha

and

c = \sqrt {1 + \sin \alpha \cos \alpha }

Clearly

a

and

b < 1

but

c > 1

as

\,\,\,\sin \alpha > 0

and

\cos \alpha > 0

$\therefore$

c

is the greatest side and greatest angle is

C

$\therefore$

\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}

= {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}

= - {1 \over 2}

$\therefore$

C = {120^ \circ }

Q42

If in a

\Delta ABC

, the altitudes from the vertices

A, B, C

on opposite sides are in H.P, then

\sin A,\sin B,\sin C

are in :

A G. P.

B A. P.

C A.P-G.P.

D H. P

Correct Answer

Option B

Solution

\Delta = {1 \over 2}{p_1}a = {1 \over 2}{p_2}b = {1 \over 2}{p_3}b

{p_1},{p_2},{p_3},

are in

H.P.

\Rightarrow {{2\Delta } \over a},{{2\Delta } \over b},{{2\Delta } \over c}

are in

H.P.

\Rightarrow {1 \over a},{1 \over b},{1 \over c},

are in

H.P.

\Rightarrow a,b,c

are in

A.P.

$\Rightarrow$

K\sin A,K\sin B,K\sin C

are in

A.P.

$\Rightarrow$

\sin A,\sin B,\sin C

are in

A.P.

Q43

In a triangle

ABC

, let

\angle C = {\pi \over 2}

. If

r

is the inradius and

R

is the circumradius of the triangle

ABC

, then

2(r+R)

equals :

A

b+c

B

a+b

C

a+b+c

D

c+a

Correct Answer

Option B

Solution

We know by sin c rule

{c \over {\sin C}} = 2R \Rightarrow c = 2R\sin C

\Rightarrow c = 2R

\left( \, \right.

as

\,\,\,\angle C = {90^ \circ }

\left. \, \right)

Also

\tan {C \over 2} = {r \over {s - c}}

\Rightarrow \tan {\pi \over 4} = {r \over {s - c}}

\left( \, \right.

as

\,\,\,\angle C = {90^ \circ }

\left. \, \right)

\Rightarrow r = s - c = {{a + b - c} \over 2}

\Rightarrow 2r + c = a + b

\Rightarrow 2r + 2R = a + b

(using

c=2R

)

Q44

In a

\Delta PQR,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu}

If

3{\mkern 1mu} \sin {\mkern 1mu} P + 4{\mkern 1mu} \cos {\mkern 1mu} Q = 6

and

4\sin Q + 3\cos P = 1,

then the angle R is equal to :

A

{{5\pi } \over 6}

B

{{\pi } \over 6}

C

{{\pi } \over 4}

D

{{3\pi } \over 4}

Correct Answer

Option B

Solution

Given

3

\sin \,P + 4\cos Q = 6

\,\,\,\,\,\,\,\,...\left( i \right)

4\sin Q + 3\cos P = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)

Squaring and adding

(i)

&

(ii)

we get

9\,{\sin ^2}P + 16{\cos ^2}Q + 24\sin P\cos Q

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 16\,{\sin ^2}Q + 9{\cos ^2}P + 24\sin Q\cos P

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

= 36 + 1 = 37

\Rightarrow 9\left( {{{\sin }^2}p + {{\cos }^2}P} \right) + 16\left( {{{\sin }^2}Q + {{\cos }^2}q} \right)

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 24\left( {\sin P\cos Q + \cos P\sin Q} \right) = 37

\Rightarrow 9 + 16 + 24\sin \left( {P + Q} \right) = 37

[ As

{\sin ^2}\theta + {\cos ^2}\theta = 1

and

\sin A\cos B + \cos A\sin B

= \sin \left( {A + B} \right)

]

\Rightarrow \sin \left( {P + Q} \right) = {1 \over 2}

\Rightarrow P + Q = {\pi \over 6}

or

{{5\pi } \over 6}

\Rightarrow R = {{5\pi } \over 6}

or

{\pi \over 6}

(as

P + Q + R = \pi

) If

R = {{5\pi } \over 6}

then

0 < P,Q < {\pi \over 6}

\Rightarrow \cos Q < 1

and

\sin P < {1 \over 2}

\Rightarrow 3\sin P + 4\cos Q < {{11} \over 2}

which is not true. So

R = {\pi \over 6}

Q45

Let the orthocentre and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the circumcentre of this triangle, then the radius of the circle having line segment AC as diameter, is :

A

{{3\sqrt 5 } \over 2}

B

\sqrt {10}

C

2\sqrt {10}

D

3\sqrt {{5 \over 2}}

Correct Answer

Option D

Solution

In a triangle, orthocentre, centroid and circumcenter are collinear and centroid divides orthocenter and circumcenter in 2 : 1 ratio.

Here AB = 2 BC

AB = \sqrt {{{\left( {3 + 3} \right)}^2} + {{\left( {3 - 5} \right)}^2}}

= \sqrt {36 + 4}

= \sqrt {40}

= 2\sqrt {10}

\therefore\,\,\,\,

BC

= \sqrt {10}

\therefore\,\,\,\,

AB + BC

= 2\sqrt {10} + \sqrt {10}

= 3\sqrt {10}

AC is the diameter of the circle.

\therefore\,\,\,

Radius

= {1 \over 2}\,\,

AC

= {1 \over 2} \times 3\sqrt {10}

= 3\sqrt {{5 \over 2}}

Q46

With the usual notation, in

\Delta

ABC, if

\angle A + \angle B

= 120o, a =

\sqrt 3

+

1, b =

\sqrt 3

-

1 then the ratio

\angle A:\angle B,

is :

A 9 : 7

B 7 : 1

C 5 : 3

D 3 : 1

Correct Answer

Option B

Solution

A + B = 120o

\tan {{A - B} \over 2} = {{a - b} \over {a + b}}\cot \left( {{c \over 2}} \right)

= {{\sqrt 3 + 1 - \sqrt 3 + 1} \over {2\left( {\sqrt 3 } \right)}}\cot \left( {{{30}^o}} \right) = {1 \over {\sqrt 3 }}.\sqrt 3 = 1

{{A - B} \over 2} = {45^o}

\Rightarrow A - B = {90^o}

\ A + B = {120^o}

2A = {210^o}

A = {105^o}

B = {15^o}

$\therefore$

\angle A:\angle B,

= 7 : 1

Q47

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :

A 5 : 9 : 13

B 5 : 6 : 7

C 4 : 5 : 6

D 3 : 4 : 5

Correct Answer

Option C

Solution

Let smallest angle

\angle A

= $\theta$ and largest angle

\angle C

is double of

\angle A

. $\therefore$

\angle C

= 2 $\theta$ $\therefore$

\angle B

= $\pi$ - 3 $\theta$ Let length of sides are a, b, c where a < b < c As a, b, c are in A.P then 2b = a + c From sin rule we can say, 2 sin B = sin A + sin C $\Rightarrow$ 2 sin ( $\pi$ – 3 $\theta$ ) = sin $\theta$ + sin 2 $\theta$ $\Rightarrow$ 2 sin (3 $\theta$ ) = sin $\theta$ + sin 2 $\theta$ $\Rightarrow$ 2(3sin $\theta$ - 4sin3 $\theta$ ) = sin $\theta$ + 2sin $\theta$ cos $\theta$ $\Rightarrow$ 2(3 - 4sin2 $\theta$ ) = 1 + 2cos $\theta$ $\Rightarrow$ 2(3 - 4 + 4cos2 $\theta$ ) = 1 + 2cos $\theta$ $\Rightarrow$ 2(4cos2 $\theta$ - 1) = 1 + 2cos $\theta$ $\Rightarrow$ 8 cos2 $\theta$ – 2 cos $\theta$ – 3 = 0 $\Rightarrow$ 8 cos2 $\theta$ – 6 cos $\theta$ + 4 cos $\theta$ – 3 = 0 $\Rightarrow$ (2 cos $\theta$ + 1) ( 4 cos $\theta$ - 3) = 0 $\Rightarrow$ cos $\theta$ =

{3 \over 4}

or cos $\theta$ =

-{1 \over 2}

But cos $\theta$ =

-{1 \over 2}

is not possible as

\angle C

is acute angle.

Now we have to find, a : b : c $\Rightarrow$ sin A : sin B : sin C [ from sin rule] $\Rightarrow$ sin $\theta$ : sin ( $\pi$ - 3 $\theta$ ) : sin 2 $\theta$ $\Rightarrow$ sin $\theta$ : sin 3 $\theta$ : sin 2 $\theta$ $\Rightarrow$ sin $\theta$ : (3sin $\theta$ - 4sin3 $\theta$ ) : 2sin $\theta$ cos $\theta$ $\Rightarrow$ 1 : (3 - 4sin2 $\theta$ ) : 2 cos $\theta$ $\Rightarrow$ 1 : (3 - 4 + 4cos2 $\theta$ ) : 2 cos $\theta$ $\Rightarrow$ 1 : (4cos2 $\theta$ - 1) : 2 cos $\theta$ $\Rightarrow$ 1 : (4

{\left( {{3 \over 4}} \right)^2}

- 1) : 2 $\times$

{{3 \over 4}}

$\Rightarrow$ 1 :

{5 \over 4}

:

{{3 \over 2}}

$\Rightarrow$ 4 : 5 : 6

Q48

The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 :

\sqrt 3

. If c = 4 cm, then the area (in sq. cm) of this triangle is :

A 2

\sqrt 3

B 4

\sqrt 3

C

{4 \over {\sqrt 3 }}

D

{2 \over {\sqrt 3 }}

Correct Answer

Option A

Solution

From the question 2B = A + C & A + B + C = $\pi$ $\Rightarrow$ 3B = $\pi$ $\Rightarrow$ B =

{\pi \over 3}

\therefore A + C = {{2\pi } \over 3}\sigma

{a \over b} = {1 \over {\sqrt 3 }}

{{2R\sin A} \over {2R\sin B}} = {1 \over {\sqrt 3 }}

sin A =

{1 \over 2}

$\therefore$ A = 30o $\therefore$ a = 2, b = 2

\sqrt 3

, c = 4

\Delta = {1 \over 2} \times 2\sqrt 3 \times 2 = 2\sqrt 3

Q49

A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If

\angle BAC = {90^o}

and area

\left( {\Delta ABC} \right) = 5\sqrt 5

s units, then the abscissa of the vertex C is :

A

1 + 2\sqrt 5

B

2\sqrt 5 - 1

C

1 + \sqrt 5

D

2 + \sqrt 5

Correct Answer

Option A

Solution

Distance between A and B

= \sqrt {{{(2)}^2} + {{( - 1)}^2}} = \sqrt 5

Area of triangle =

5\sqrt 5

\Rightarrow {1 \over 2} \times \sqrt 5 \times x = 5\sqrt 5

\Rightarrow x = 10

Slope of AB,

{m_{AB}} = {{1 - 2} \over {3 - 1}} = - {1 \over 2}

AC is perpendicular to AB. $\therefore$

{m_{AB}}\,.\,{m_{AC}} = - 1

\Rightarrow {m_{AC}} = 2 = \tan \theta

So,

\sin \theta = {2 \over {\sqrt 5 }}

,

\cos \theta = {1 \over {\sqrt 5 }}

$\therefore$

a = {x_A} + r\cos \theta

= 1 + 10 \times {1 \over {\sqrt 5 }}

= 1 + 2\sqrt 5

Q50

Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be

\left( {{{10} \over 3},{7 \over 3}} \right)

. If

\alpha

,

\beta

are the roots of the equation

a{x^2} + bx + 1 = 0

, then the value of

{\alpha ^2} + {\beta ^2} - \alpha \beta

is :

A

{{69} \over {256}}

B

{{71} \over {256}}

C

- {{71} \over {256}}

D

- {{69} \over {256}}

Correct Answer

Option C

Solution

2b = a + c

{{2a + 2} \over 3} = {{10} \over 3}

and

{{2b + c} \over 3} = {7 \over 3}

a = 4,

\left\{ \begin{array}{ll}2b + c = 7 \\ 2b - c = 4 \end{array} \right\}

, solving

b = {{11} \over 4}

c = {3 \over 2}

$\therefore$ Quadratic Equation is

4{x^2} + {{11} \over 4}x + 1 = 0

$\therefore$ The value of

{\alpha ^2} + {\beta ^2} - \alpha \beta

=

{\alpha ^2} + {\beta ^2} + 2\alpha \beta - 3\alpha \beta

=

{(\alpha + \beta )^2} - 3\alpha \beta

= {{121} \over {256}} - {3 \over 4} = - {{71} \over {256}}